Solving First order linear differential equation

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  • Опубликовано: 6 янв 2025

Комментарии • 14

  • @nihil1312
    @nihil1312 Год назад +7

    Love this professor. Never stop teaching, because those who stop teaching, stop living.

    • @PrimeNewtons
      @PrimeNewtons  Год назад +4

      🤣 never heard that before......

  • @EliškaPangrácová
    @EliškaPangrácová Месяц назад +1

    You are awesome, I really love your energy. I was so depressed from going through the books tryint to understand it….but this video made me happier about maths and I get the integrating factor, thank you, ❤

  • @mishradevansh308
    @mishradevansh308 Год назад +3

    Salute to your consistency! 🔥🔥

  • @Legend-nw1jh
    @Legend-nw1jh 5 месяцев назад

    I feel myself vey lucky my 12th is going on and i got your channel

  • @peterchege4616
    @peterchege4616 Год назад +2

    Could you do videos on 2nd order PDEs i.e homogenous and non-homogenous PDEs as well as the solution of the wave equation

  • @francaisdeuxbaguetteiii7316
    @francaisdeuxbaguetteiii7316 Год назад +3

    9:25 hey! I love your videos, im currently bingewatching them all! The calm voice and amazing descriptions are simply the best. I just have a note to say here! Integrating d(xy) would still yield a constant. We just typically dont write it because if you think about it,
    f(x)+d = g(x)+c
    Is the same as
    f(x)= g(x)+c-d
    c-d is still a constant so we usually just shorten this step and set the “c-d” to just be +c on the right hand side, so technically your justification for not writing +d on the left is incorrect. Still doesnt change the meaning and educational value of this lesson, and keep making videos! ❤

  • @kingbeauregard
    @kingbeauregard Год назад +4

    Pay close attention to what's going on at 5:20; that's the secret to the entire problem. The Integrating Factor is applied so that the equation becomes an "exact" equation -- in other words, the left side of the equation becomes (xy)'.
    In general, the Integrating factor satisfies the following requirement: what u(x) can you create such that u'(x) = u(x)*P(x)? Well, that would be u(x) = e^(integral of P(x)); when you differentiate it you get u'(x) = u(x)*P(x).

  • @engrlois
    @engrlois 3 месяца назад +1

    September 9, 2024 @11:54 pm.
    2nd year, DE.

    • @engrlois
      @engrlois 3 месяца назад

      Mandatory leaving comments sa lahat ng videos na nasolve ko hehe, thank you so much!!!